Question: Simplify the following expression: $y = \dfrac{3x^2- 11x- 20}{x - 5}$
Answer: First use factoring by grouping to factor the expression in the numerator. This expression is in the form ${A}x^2 + {B}x + {C}$ First, find two values, $a$ and $b$ , so: $ \begin{eqnarray} {ab} &=& {A}{C} \\ {a} + {b} &=& {B} \end{eqnarray} $ In this case: $ \begin{eqnarray} {ab} &=& {(3)}{(-20)} &=& -60 \\ {a} + {b} &=& &=& {-11} \end{eqnarray} $ In order to find ${a}$ and ${b}$ , list out the factors of $-60$ and add them together. Remember, since $-60$ is negative, one of the factors must be negative. The factors that add up to ${-11}$ will be your ${a}$ and ${b}$ When ${a}$ is ${4}$ and ${b}$ is ${-15}$ $ \begin{eqnarray} {ab} &=& ({4})({-15}) &=& -60 \\ {a} + {b} &=& {4} + {-15} &=& -11 \end{eqnarray} $ Next, rewrite the expression as $({A}x^2 + {a}x) + ({b}x + {C})$ $ ({3}x^2 +{4}x) + ({-15}x {-20}) $ Factor out the common factors: $ x(3x + 4) - 5(3x + 4)$ Now factor out $(3x + 4)$ $ (3x + 4)(x - 5)$ The original expression can therefore be written: $ \dfrac{(3x + 4)(x - 5)}{x - 5}$ We are dividing by $x - 5$ , so $x - 5 \neq 0$ Therefore, $x \neq 5$ This leaves us with $3x + 4; x \neq 5$.